Question

Determine the tangential and centripetal components of the net force exerted on the car (by the ground) in Example 5–8 when its speed is 15 m/s. The car’s mass is 950 kg.

Short answer

The tangential component of the net force exerted on the car is $3.0\times {10}^3\mathrm{N}$and the centripetal component of the net force exerted on the car is $4.3\times {10}^2\mathrm{N}$.

Step-by-step solution

Step 1. Understanding the centripetal and tangential acceleration

When the net force exerted on the car of mass m moving in a circular path of radius r having a constant speed v does not act towards the center of the circle but acts at an angle, then the motion of car is termed as non-uniform circular motion.

This force has two components, tangential component which acts along the tangent and centripetal component which acts along the centre of the circle.

The magnitude of the centripetal acceleration is given as:

$a_{\mathrm{R}}=\frac{v^2}{r}$

The magnitude of the tangential acceleration is given as:

$a_{\mathrm{t}}=\frac{\Delta v}{\Delta t}$

Step 2. Identification of the given information

• The speed of the car is, v = 15 m/s.
• The mass of the car is, m = 950 kg.
• The radius of the circular track is, r = 500 m.
• The tangential acceleration is, $a_{\mathrm{t}}=3.2\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Determination of the tangential component of the net force

The tangential component of the net force is given as:

$F_{\mathrm{t}}=m{a}_{\mathrm{t}}$

Substitute all the values in the above equation.

$\begin{array}{c}{F}_{\mathrm{t}}=\left(950\mathrm{kg}\right)\left(3.2\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =3040\mathrm{N}\\ \approx 3.0\times {10}^3\mathrm{N}\end{array}$

Thus, the tangential component of the net force exerted on the car is $3.0\times {10}^3\mathrm{N}$.

Step 4. Determination of the centripetal force component of the net force

The centripetal component of the net force is given as:

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

Substitute all the values in the above equation.

$\begin{array}{c}{F}_{\mathrm{R}}=\frac{\left(950\mathrm{kg}\right){\left(15\mathrm{m}/\mathrm{s}\right)}^2}{\left(500\mathrm{m}\right)}\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =427.5\mathrm{N}\\ \approx 4.3\times {10}^2\mathrm{N}\end{array}$

Thus, the centripetal component of the net force exerted on the car is $4.3\times {10}^2\mathrm{N}$.